Since the time of interest is very much less than the circuit time constant the method used is Ok. One has to be careful when this is not the case. When the time of interest is say 1% or more of the circuit time constant then a more thorough analysis would be in order if one needed to be precise.

Say I have a series DC circuit with a 100V DC power supply, a switch, a 300mH inductor and a 10 Ohm resistor.

If I close the switch for 1uS how can I calculate the current through the inductor at that time?

Initially I thought of using V*Ton/L=di/dt to get 100*.000001 / .3 = 333uA @ 1uS. Is this right?

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Check your units:

V*Ton/L=di/dt

You have Vs/H (volt-seconds per henry) on the left and A/s (amperes per second) on the right. Are those consistent? Keep in mind that 1H=1Vs/A.

Your symbolic equation and your numeric equation are not consistent with each other. Your numeric equation, with units, is

I(1μs) = 100V*1μs/300mH

This implies that your governing equation is

I(t) = V*t/L

Does this make sense? It says that the current will increase linearly with time.

And notice that you are also claiming that resistor has no effect on the circuit. Does this make sense.

Note that the answer to these two questions is not entirely trivial.

You are making the classic mistake of "I have an equation that needs something plugged in that is a time and, by the way, I have a parameter that is a time. So I must need to throw that parameter into the equation."

In this case, and I fear it is by coincidence and not by careful consideration, this mistake yields a result that, for a short period of time, provides a reasonable approximation to the actual answer.

To check this, what answer would you have come up with if asked to find the current at 1s instead of 1μs?